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Dual Nature of Matter and Radiation: Physics | JEE Main

Q: Does the charge of the particle matter for the de Broglie wavelength?

No, because the kinetic energy is directly given. Charge is only relevant if kinetic energy was acquired by accelerating through an electric potential ( ).

The kinetic energy of an electron,

\alpha

-particle and a proton are given as

4\mathrm{K}

,

2\mathrm{K}

and

\mathrm{K}

respectively. The de-Broglie wavelength associated with electron

(\lambda\mathrm{e})

,

\alpha

-particle

(\lambda\alpha)

and the proton

(\lambda\mathrm{p})

are as follows :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

\lambda\alpha = \lambda\mathrm{p} < \lambda\mathrm{e}

B

\lambda\alpha > \lambda\mathrm{p} > \lambda\mathrm{e}

C

\lambda\alpha < \lambda\mathrm{p} < \lambda\mathrm{e}

D

\lambda\alpha = \lambda\mathrm{p} > \lambda\mathrm{e}

Step-by-step solutionView

Correct answer

Because $\lambda = \frac{h}{\sqrt{2mK}}$ , the product $mK$ is largest for the $\alpha$ -particle and smallest for the electron, making $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$ .

Option analysis

Why each option works or fails

A ·

\lambda\alpha = \lambda\mathrm{p} < \lambda\mathrm{e}

Believing that $m_\alpha K_\alpha = m_{\mathrm{p}} K_{\mathrm{p}}$ by assuming the mass ratio between $\alpha$ and proton is $2:1$ instead of $4:1$ . Recall that an $\alpha$ -particle consists of 2 protons and 2 neutrons, so $m_\alpha \approx 4 m_{\mathrm{p}}$ , giving $m_\alpha K_\alpha = 4 m_{\mathrm{p}} \times 2K = 8 m_{\mathrm{p}} K \neq m_{\mathrm{p}} K$ .

B ·

\lambda\alpha > \lambda\mathrm{p} > \lambda\mathrm{e}

Assuming that de Broglie wavelength is directly proportional to momentum or $\sqrt{mK}$ rather than inversely proportional. Use the de Broglie relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$ , which shows an inverse relationship with $\sqrt{mK}$ .

C ·

\lambda\alpha < \lambda\mathrm{p} < \lambda\mathrm{e}

None. This is the correct relationship. Comparing the values of $mK$ : for electron $m_{\mathrm{e}}(4K)$ , for proton $m_{\mathrm{p}}(K)$ , and for $\alpha$ -particle $(4m_{\mathrm{p}})(2K) = 8m_{\mathrm{p}}K$ . Since $m_{\mathrm{e}} \ll m_{\mathrm{p}}$ , $(mK)_\alpha > (mK)_{\mathrm{p}} > (mK)_{\mathrm{e}}$ , which gives $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$ .

D ·

\lambda\alpha = \lambda\mathrm{p} > \lambda\mathrm{e}

Confusing mass ratios and inverting the proportionality between $\lambda$ and $\sqrt{mK}$ . Ensure both the correct inverse proportionality $\lambda \propto 1/\sqrt{mK}$ and the correct mass ratio $m_\alpha \approx 4m_{\mathrm{p}}$ are applied.

Reviewed route

Solution

StepWorking

01given

Kinetic energies: $K_e = 4K$ , $K_\alpha = 2K$ , $K_p = K$ . Relative masses: $m_p = m$ , $m_\alpha = 4m$ , $m_e \approx m/1840 \ll m$ .

02strategise

de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$ . Therefore, $\lambda \propto \frac{1}{\sqrt{m K}}$ . Compare the product $(m \cdot K)$ for each particle.

03execute

Calculate the effective $(m \cdot K)$ factors: For electron: $(m_e K_e) = \frac{m}{1840} \times 4K = \frac{mK}{460}$ For proton: $(m_p K_p) = m \times K = mK$ For $\alpha$ -particle: $(m_\alpha K_\alpha) = 4m \times 2K = 8mK$ Comparing products: $(m_e K_e) < (m_p K_p) < (m_\alpha K_\alpha)$ Since $\lambda \propto \frac{1}{\sqrt{mK}}$ , we get: $\lambda_\alpha < \lambda_p < \lambda_e$

✓verify

Since mass of electron is drastically smaller than that of nucleons, its momentum is much smaller despite higher kinetic energy, leading to the largest wavelength. Alpha particle has the largest mass and higher KE than proton, so its momentum is highest and wavelength is smallest.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Does the charge of the particle matter for the de Broglie wavelength?

No, because the kinetic energy is directly given. Charge is only relevant if kinetic energy was acquired by accelerating through an electric potential ( $K = qV$ ).

Answer

Because $\lambda = \frac{h}{\sqrt{2mK}}$ , the product $mK$ is largest for the $\alpha$ -particle and smallest for the electron, making $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$ .

Why each option works or fails

A: $\lambda\alpha = \lambda\mathrm{p} < \lambda\mathrm{e}$ - Believing that $m_\alpha K_\alpha = m_{\mathrm{p}} K_{\mathrm{p}}$ by assuming the mass ratio between $\alpha$ and proton is $2:1$ instead of $4:1$ . Recall that an $\alpha$ -particle consists of 2 protons and 2 neutrons, so $m_\alpha \approx 4 m_{\mathrm{p}}$ , giving $m_\alpha K_\alpha = 4 m_{\mathrm{p}} \times 2K = 8 m_{\mathrm{p}} K \neq m_{\mathrm{p}} K$ .
B: $\lambda\alpha > \lambda\mathrm{p} > \lambda\mathrm{e}$ - Assuming that de Broglie wavelength is directly proportional to momentum or $\sqrt{mK}$ rather than inversely proportional. Use the de Broglie relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$ , which shows an inverse relationship with $\sqrt{mK}$ .
C · correct: $\lambda\alpha < \lambda\mathrm{p} < \lambda\mathrm{e}$ - None. This is the correct relationship. Comparing the values of $mK$ : for electron $m_{\mathrm{e}}(4K)$ , for proton $m_{\mathrm{p}}(K)$ , and for $\alpha$ -particle $(4m_{\mathrm{p}})(2K) = 8m_{\mathrm{p}}K$ . Since $m_{\mathrm{e}} \ll m_{\mathrm{p}}$ , $(mK)_\alpha > (mK)_{\mathrm{p}} > (mK)_{\mathrm{e}}$ , which gives $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$ .
D: $\lambda\alpha = \lambda\mathrm{p} > \lambda\mathrm{e}$ - Confusing mass ratios and inverting the proportionality between $\lambda$ and $\sqrt{mK}$ . Ensure both the correct inverse proportionality $\lambda \propto 1/\sqrt{mK}$ and the correct mass ratio $m_\alpha \approx 4m_{\mathrm{p}}$ are applied.

Step-by-step solution

given: Kinetic energies: $K_e = 4K$ , $K_\alpha = 2K$ , $K_p = K$ . Relative masses: $m_p = m$ , $m_\alpha = 4m$ , $m_e \approx m/1840 \ll m$ .
strategise: de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$ . Therefore, $\lambda \propto \frac{1}{\sqrt{m K}}$ . Compare the product $(m \cdot K)$ for each particle.
execute: Calculate the effective $(m \cdot K)$ factors: For electron: $(m_e K_e) = \frac{m}{1840} \times 4K = \frac{mK}{460}$ For proton: $(m_p K_p) = m \times K = mK$ For $\alpha$ -particle: $(m_\alpha K_\alpha) = 4m \times 2K = 8mK$ Comparing products: $(m_e K_e) < (m_p K_p) < (m_\alpha K_\alpha)$ Since $\lambda \propto \frac{1}{\sqrt{mK}}$ , we get: $\lambda_\alpha < \lambda_p < \lambda_e$
verify: Since mass of electron is drastically smaller than that of nucleons, its momentum is much smaller despite higher kinetic energy, leading to the largest wavelength. Alpha particle has the largest mass and higher KE than proton, so its momentum is highest and wavelength is smallest.

Shortcut:

When to use it: Direct inspection using lambda ~ 1/sqrt(m*K)

quick_kill: Using $\lambda \propto 1/\sqrt{mK}$ : $mK$ values are $\approx 0$ (electron), $1$ (proton), $8$ (alpha). Hence, $\lambda_\alpha < \lambda_p < \lambda_e$ immediately gives Option (2).

Dual Nature of Matter and Radiation: Physics | JEE Main

What feels right?

Because $\lambda = \frac{h}{\sqrt{2mK}}$ , the product $mK$ is largest for the $\alpha$ -particle and smallest for the electron, making $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$ .

Why each option works or fails

Solution

01given

02strategise

03execute

✓verify

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